Question Number 54374 by maxmathsup by imad last updated on 02/Feb/19
$${calculate}\:{h}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({at}\right)}{{ch}\left(\frac{{t}}{\mathrm{2}}\right)}{dt}\:. \\ $$
Commented by Abdo msup. last updated on 08/Feb/19
![h(a)=_((t/2)=x) 2 ∫_(−∞) ^(+∞) ((cos(2ax))/(ch(x)))dx =4 ∫_0 ^∞ ((cos(2ax))/((e^x +e^(−x) )/2))dx =8 ∫_0 ^∞ ((e^(−x) cos(2ax))/(1+e^(−2x) ))dx =8 ∫_0 ^∞ e^(−x) cos(2ax)(Σ_(n=0) ^∞ (−1)^n e^(−2nx) )dx =8 Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(2n+1)x) cos(2ax)dx let A_n =∫_0 ^∞ e^(−(2n+1)x) cos(2ax)dx ⇒ A_n =Re( ∫_0 ^∞ e^(−(2n+1)x) e^(2iax) dx) but ∫_0 ^∞ e^((−(2n+1)+2ia)x) dx=[(1/(−(2n+1)+2ia)) e^((−(2n+1)+2ia)x) ]_0 ^(+∞) = (1/(−(2n+1)+2ia)) =−(1/((2n+1)−2ia)) =−((2n+1 +2ia)/((2n+1)^2 +4a^2 )) ⇒ A_n =−((2n+1)/((2n+1)^2 +4a^2 )) ⇒ h(a)=8 Σ_(n=0) ^∞ (−1)^(n+1) ((2n+1)/((2n+1)^2 +4a^2 )) and this serie can be calculated by fourier series ...be continued...](https://www.tinkutara.com/question/Q54638.png)
$${h}\left({a}\right)=_{\frac{{t}}{\mathrm{2}}={x}} \:\:\:\mathrm{2}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{ax}\right)}{{ch}\left({x}\right)}{dx} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{ax}\right)}{\frac{{e}^{{x}} \:+{e}^{−{x}} }{\mathrm{2}}}{dx}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} {cos}\left(\mathrm{2}{ax}\right)}{\mathrm{1}+{e}^{−\mathrm{2}{x}} }{dx} \\ $$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} {cos}\left(\mathrm{2}{ax}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{e}^{−\mathrm{2}{nx}} \right){dx} \\ $$$$=\mathrm{8}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} \:{cos}\left(\mathrm{2}{ax}\right){dx} \\ $$$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} {cos}\left(\mathrm{2}{ax}\right){dx}\:\Rightarrow \\ $$$${A}_{{n}} ={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} {e}^{\mathrm{2}{iax}} {dx}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}{ia}\right){x}} {dx}=\left[\frac{\mathrm{1}}{−\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}{ia}}\:{e}^{\left(−\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}{ia}\right){x}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\:\frac{\mathrm{1}}{−\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{2}{ia}}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{2}{ia}} \\ $$$$=−\frac{\mathrm{2}{n}+\mathrm{1}\:+\mathrm{2}{ia}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}{a}^{\mathrm{2}} }\:\Rightarrow\:{A}_{{n}} =−\frac{\mathrm{2}{n}+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}{a}^{\mathrm{2}} }\:\Rightarrow \\ $$$${h}\left({a}\right)=\mathrm{8}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:\frac{\mathrm{2}{n}+\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${and}\:{this}\:{serie}\:{can}\:{be}\:{calculated}\:{by}\:\:{fourier}\:{series} \\ $$$$…{be}\:{continued}… \\ $$